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Sagot :
To find the value of [tex]\(x\)[/tex] that would make [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex], we can use the side-splitter theorem. According to the theorem, if [tex]\(\overline{KM}\)[/tex] is parallel to [tex]\(\overline{JN}\)[/tex], then the ratio [tex]\( \frac{JK}{KL} \)[/tex] must be equal to the ratio [tex]\( \frac{KM}{MN} \)[/tex].
Given the problem:
[tex]\[ \frac{x-5}{x}=\frac{x-3}{x+4} \][/tex]
Follow these steps to solve for [tex]\(x\)[/tex]:
1. Cross-multiply the given proportion:
[tex]\[ (x-5)(x+4) = x(x-3) \][/tex]
2. Distribute the terms:
[tex]\[ x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3) \][/tex]
[tex]\[ x^2 + 4x - 5x - 20 = x^2 - 3x \][/tex]
3. Simplify the equation:
[tex]\[ x^2 - x - 20 = x^2 - 3x \][/tex]
4. To eliminate [tex]\(x^2\)[/tex] from both sides, subtract [tex]\(x^2\)[/tex] from both sides of the equation:
[tex]\[ -x - 20 = -3x \][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[ -x - 20 = -3x \][/tex]
[tex]\[ -20 = -3x + x \][/tex]
[tex]\[ -20 = -2x \][/tex]
[tex]\[ x = \frac{20}{2} \][/tex]
[tex]\[ x = 10 \][/tex]
So, the value of [tex]\(x\)[/tex] that makes [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex] is [tex]\(x = 10\)[/tex]:
Summarizing the steps with the final solution:
- By the converse of the side-splitter theorem, if [tex]\( \frac{JK}{KL} = \frac{KM}{MN} \)[/tex]:
[tex]\[ \frac{x-5}{x} = \frac{x-3}{x+4} \][/tex]
- Substitute the expressions into the proportion:
[tex]\[ \frac{x-5}{x} = \frac{x-3}{x+4} \][/tex]
- Cross-multiply:
[tex]\[ (x-5)(x+4) = x(x-3) \][/tex]
- Distribute terms:
[tex]\[ x^2 - x - 20 = x^2 - 3x \][/tex]
- Simplify and solve for [tex]\(x\)[/tex]:
[tex]\[ x = 10 \][/tex]
Therefore, [tex]\( x = 10 \)[/tex] makes [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex].
Given the problem:
[tex]\[ \frac{x-5}{x}=\frac{x-3}{x+4} \][/tex]
Follow these steps to solve for [tex]\(x\)[/tex]:
1. Cross-multiply the given proportion:
[tex]\[ (x-5)(x+4) = x(x-3) \][/tex]
2. Distribute the terms:
[tex]\[ x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3) \][/tex]
[tex]\[ x^2 + 4x - 5x - 20 = x^2 - 3x \][/tex]
3. Simplify the equation:
[tex]\[ x^2 - x - 20 = x^2 - 3x \][/tex]
4. To eliminate [tex]\(x^2\)[/tex] from both sides, subtract [tex]\(x^2\)[/tex] from both sides of the equation:
[tex]\[ -x - 20 = -3x \][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[ -x - 20 = -3x \][/tex]
[tex]\[ -20 = -3x + x \][/tex]
[tex]\[ -20 = -2x \][/tex]
[tex]\[ x = \frac{20}{2} \][/tex]
[tex]\[ x = 10 \][/tex]
So, the value of [tex]\(x\)[/tex] that makes [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex] is [tex]\(x = 10\)[/tex]:
Summarizing the steps with the final solution:
- By the converse of the side-splitter theorem, if [tex]\( \frac{JK}{KL} = \frac{KM}{MN} \)[/tex]:
[tex]\[ \frac{x-5}{x} = \frac{x-3}{x+4} \][/tex]
- Substitute the expressions into the proportion:
[tex]\[ \frac{x-5}{x} = \frac{x-3}{x+4} \][/tex]
- Cross-multiply:
[tex]\[ (x-5)(x+4) = x(x-3) \][/tex]
- Distribute terms:
[tex]\[ x^2 - x - 20 = x^2 - 3x \][/tex]
- Simplify and solve for [tex]\(x\)[/tex]:
[tex]\[ x = 10 \][/tex]
Therefore, [tex]\( x = 10 \)[/tex] makes [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex].
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